# Nuclear Decay

1. Sep 15, 2008

### iamalexalright

1. The problem statement, all variables and given/known data

In a sample of one litre of carbon dioxide at STP an average of 5 disintegrations

C -> N + e + v

are observed per minute. Calculate the atomic fraction of C present if the mean lifetime of this nucleus is 8267 years.

2. Relevant equations

$$\tau$$ = $$\omega$$-1 = 8267 years

N(t) = N0 * $$e$$-$$(\omega$$ * t)

3. The attempt at a solution

First I found $$\tau$$ = 2.607 * 1011 seconds.

And really I'm at a loss of where to go next. I figure N(t)/N0 is what im looking for but I cannot solve for this without knowing the time t and I do not know where the 5 disintegrations per minute comes in. I'm not looking for an answer(already know it), i just want a little push in the right direction.

2. Sep 15, 2008

### gabbagabbahey

Since there are an average of 5 disintegrations per minute, you would expect:

$$N(t)-N(t+60s)=5$$...can you take it from there?

3. Sep 15, 2008

### Staff: Mentor

One can find the total amount of carbon from "one litre of carbon dioxide" - simply find the moles of CO2 in 1 liter at STP. It would appear that one is to find the atomic fraction of radiocarbon or C-14.

The activity, decays per unit time, is the product of the decay constant $\lambda$ and the number of atoms of the radionuclide present, i.e. A(t) = $\lambda$ N(t). One is given A, so find N.